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Math  /  Discrete

QuestionA small farm has five goats, six horses, seven cows, and eight pigs. The farmers would like to line up two animals of each species for a photograph. How many different pictures are possible if each species of animal stands together (that is, the two goats stand side-by-side, and the two horses stand side-by-side, etc.)?
The number of possible pictures is □\square

Studdy Solution

STEP 1

1. There are five goats, six horses, seven cows, and eight pigs.
2. We need to select two animals of each species and line them up for a photograph.
3. Each species' pair stands together in the lineup.
4. The order of species pairs in the lineup matters.

STEP 2

1. Determine the number of ways to choose two animals from each species.
2. Determine the number of ways to arrange the chosen pairs in a lineup.
3. Calculate the total number of different pictures possible.

STEP 3

Calculate the number of ways to choose 2 goats from 5 goats. This is given by the combination formula:
(52)=5×42×1=10\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10
Calculate the number of ways to choose 2 horses from 6 horses:
(62)=6×52×1=15\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15
Calculate the number of ways to choose 2 cows from 7 cows:
(72)=7×62×1=21\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21
Calculate the number of ways to choose 2 pigs from 8 pigs:
(82)=8×72×1=28\binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28

STEP 4

Each species pair can be arranged in 2! ways. So, calculate the number of ways to arrange each pair:
2!=22! = 2

STEP 5

Calculate the number of ways to arrange the four species pairs in a lineup. There are 4 pairs, so:
4!=244! = 24

STEP 6

Calculate the total number of different pictures possible by multiplying the number of ways to choose and arrange the animals:
(10×2)×(15×2)×(21×2)×(28×2)×24(10 \times 2) \times (15 \times 2) \times (21 \times 2) \times (28 \times 2) \times 24
=20×30×42×56×24= 20 \times 30 \times 42 \times 56 \times 24
Calculate step-by-step:
1. 20×30=600 20 \times 30 = 600
2. 600×42=25200 600 \times 42 = 25200
3. 25200×56=1411200 25200 \times 56 = 1411200
4. 1411200×24=33868800 1411200 \times 24 = 33868800
The number of possible pictures is:
33868800\boxed{33868800}

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